Cuda GPUassert:遇到非法内存访问
Cuda GPUassert:遇到非法内存访问
提问于 2020-12-14 17:47:05
我试图使用__device __变量来制作游戏程序，而不是使用cudaMalloc动态声明它，但它一直告诉我，在调用cudaDeviceSynchronization()的最后一行中遇到了GPUassert:非法内存访问。我已经尝试过使用cudaMalloc的版本，而且效果很好。
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <iostream>
#include <cmath>
#include <stdio.h>
#include <stdlib.h>
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char* file, int line, bool abort = true)
{
if (code != cudaSuccess)
{
fprintf(stderr, "GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
#define M 3
#define N 3
#define K 3
using namespace std;
__device__ double A_dev[M * K];
__device__ double B_dev[K * N];
__device__ double C_dev[M * N];
__global__ void gemm(double* A, double* B, double* C, int m, int n, int k)
{
int x = blockDim.x * blockIdx.x + threadIdx.x;
int y = blockDim.y * blockIdx.y + threadIdx.y;
int i = x * n + y;
double sum = 0.0;
for (int j = 0; j < k; j++)
{
sum += A[x * k + j] * B[n * j + y];
}
C[i] = sum;
printf("The value is %f", C[i]);
}
int main(void)
{
double A_h[M * K];
double B_h[K * N];
double C_h[M * N];
for (int i = 0; i < M*K; i++)
{
A_h[i] = (double)i;
B_h[i] = (double)i;
C_h[i] = 0.0;
}
gpuErrchk(cudaMemcpyToSymbol(A_dev, A_h, M * K * sizeof(double), 0, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpyToSymbol(B_dev, B_h, K * N * sizeof(double), 0, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpyToSymbol(C_dev, C_h, M * N * sizeof(double), 0, cudaMemcpyHostToDevice));
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
dim3 dimGrid(1, 1, 1);
dim3 dimBlock(3, 3, 1);
gemm <<<dimGrid, dimBlock >>> (A_dev, B_dev, C_dev, 3, 3, 3);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
gpuErrchk(cudaMemcpyFromSymbol(C_h, C_dev, M * N * sizeof(double), 0, cudaMemcpyDeviceToHost));
return 0;
}回答 1
Stack Overflow用户
回答已采纳
发布于 2020-12-14 18:00:41
当使用__device__变量时，它们本质上处于全局范围，我们不将这些变量作为内核参数传递。您可以在内核代码中直接使用这些变量，而不必为它们设置内核参数。
如果您对代码进行了以下更改，它将无错误地运行：
#include <iostream>
#include <cmath>
#include <stdio.h>
#include <stdlib.h>
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char* file, int line, bool abort = true)
{
if (code != cudaSuccess)
{
fprintf(stderr, "GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
#define M 3
#define N 3
#define K 3
using namespace std;
__device__ double A_dev[M * K];
__device__ double B_dev[K * N];
__device__ double C_dev[M * N];
__global__ void gemm(int m, int n, int k)
{
int x = blockDim.x * blockIdx.x + threadIdx.x;
int y = blockDim.y * blockIdx.y + threadIdx.y;
int i = x * n + y;
double sum = 0.0;
for (int j = 0; j < k; j++)
{
sum += A_dev[x * k + j] * B_dev[n * j + y];
}
C_dev[i] = sum;
printf("The value is %f", C_dev[i]);
}
int main(void)
{
double A_h[M * K];
double B_h[K * N];
double C_h[M * N];
for (int i = 0; i < M*K; i++)
{
A_h[i] = (double)i;
B_h[i] = (double)i;
C_h[i] = 0.0;
}
gpuErrchk(cudaMemcpyToSymbol(A_dev, A_h, M * K * sizeof(double), 0, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpyToSymbol(B_dev, B_h, K * N * sizeof(double), 0, cudaMemcpyHostToDevice));
gpuErrchk(cudaMemcpyToSymbol(C_dev, C_h, M * N * sizeof(double), 0, cudaMemcpyHostToDevice));
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
dim3 dimGrid(1, 1, 1);
dim3 dimBlock(3, 3, 1);
gemm <<<dimGrid, dimBlock >>> (3, 3, 3);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaDeviceSynchronize());
gpuErrchk(cudaMemcpyFromSymbol(C_h, C_dev, M * N * sizeof(double), 0, cudaMemcpyDeviceToHost));
return 0;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/65293876
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